Let $E \subset \mathbb{R}^m$, $F \subset \mathbb{R}^n$ and $\mu$ the Lebesgue measure.
How to prove:
If $E$ is a Lebesgue null set of $\mathbb{R}^m \Rightarrow E \times F$ is a Lebesgue null set of $\mathbb{R}^{m+n}$.
I know there is a way to show that implication with the theorems of Fubini, Tonelli and Cavalieri, but how can it be proved without them?
I used:
Since $E \times F$ is a rectangle, by definition: $\mu_{m+n}(E \times F)=\mu_m(E) \cdot \mu_n(F)$.
So if $\mu_m(E)=0\Rightarrow \mu_{m+n}(E \times F)=0$
Is this way correct? And how can it be shown if $\mu_n(F)=\infty$?
Hint: consider $F_k = F \cap \prod_{i=1}^n [-k,k]$. From you definition of product measure, $\mu(E\times F_k)=0$. Apply the regularity of measure and let $k\to+\infty$ to conclude.